H(t)=16t2+96t+112

Solution for H(t)=16t2+96t+112 equation:

(H)=16H^2+96H+112

We move all terms to the left:

(H)-(16H^2+96H+112)=0

We get rid of parentheses

-16H^2+H-96H-112=0

We add all the numbers together, and all the variables

-16H^2-95H-112=0

a = -16; b = -95; c = -112;
Δ = b2-4ac
Δ = -952-4·(-16)·(-112)
Δ = 1857
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-95)-\sqrt{1857}}{2*-16}=\frac{95-\sqrt{1857}}{-32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-95)+\sqrt{1857}}{2*-16}=\frac{95+\sqrt{1857}}{-32}
$